The annihilator ideal graph of a commutative ring
Mojgan Afkhami
Department of Mathematics, University of Neyshabur, P.O.Box 91136-899, Neyshabur, Iran
mojgan.afkhami@yahoo.com
Nesa Hoseini
Department of Pure Mathematics, Ferdowsi University of Mashhad, P.O.Box 1159-91775, Mashhad, Iran
nesa.hoseini@gmail.com
Kazem Khashyarmanesh
Department of Pure Mathematics, Ferdowsi University of Mashhad, P.O.Box 1159-91775, Mashhad, Iran
khashyar@ipm.ir
Received: 31.10.2014; accepted: 23.3.2015.
Abstract. Let R be a commutative ring with nonzero identity and I be a proper ideal of R. The annihilator graph of R with respect to I, which is denoted by AGI(R), is the undirected graph with vertex-set V (AGI(R)) = {x ∈ R \ I : xy ∈ I for some y /∈ I} and two distinct vertices x and y are adjacent if and only if AI(xy) 6= AI(x) ∪ AI(y), where AI(x) = {r ∈ R : rx ∈ I}. In this paper, we study some basic properties of AGI(R), and we characterise when AGI(R) is planar, outerplanar or a ring graph. Also, we study the graph AGI(Zn), where Znis the ring of integers modulo n.
Keywords: Zero-divisor graph, Annihilator graph, Girth, Planar graph, Outerplanar, Ring graph
MSC 2000 classification:primary 05C10, 05C99, secondary 13A99
Introduction
Let R be a commutative ring with nonzero identity and let Z(R) be the set of zero-divisors of R. Recently, there has been considerable attention in the litera- ture to associating graphs with algebraic structures (see [1, 9, 10, 11, 12]). Prob- ably the most attention has been to the zero-divisor graph, which is denoted by Γ(R). The set of vertices of Γ(R) is Z(R)∗= Z(R)\{0}, and two distinct vertices x and y are adjacent if and only if xy = 0. The concept of the zero-divisor graph goes back to Beck [5], who let all elements of R be vertices and was mainly inter- ested in colorings. The zero-divisor graph was introduced and studied by D. F.
http://siba-ese.unisalento.it/ c 2016 Universit`a del Salento
Anderson and P. S. Livingston in [2]. For a recent survey article on zero-divisor graphs see [3]. In [4], A. Badawi introduced the annihilator graph AG(R) for a commutative ring R. Let a ∈ R and annR(a) = {r ∈ R : ra = 0}. The annihila- tor graph of R is the (undirected) graph with vertices Z(R)∗, and two distinct vertices x and y are adjacent if and only if annR(xy) 6= annR(x) ∪ annR(y).
Clearly, each edge of Γ(R) is an edge of AG(R), and so Γ(R) is a subgraph of AG(R).
In [13], S. P. Redmond introduced the zero-divisor graph of a commutative ring R with respect to an ideal I of R. Let I be a proper ideal of R. The zero-divisor graph of R with respect to I, denoted by ΓI(R), is the graph whose vertices are the set
{x ∈ R \ I : xy ∈ I for some y /∈ I},
and two distinct vertices x and y are adjacent if and only if xy ∈ I. In [13], among other results the relationship between the graphs ΓI(R) and Γ(R/I) was explored.
In this article, we introduce the annihilator graph of R with respect to a proper ideal I of R, which is denoted by AGI(R). Let x ∈ R and AI(x) = {r ∈ R : rx ∈ I}. The annihilator graph of R with respect to I is the (undirected) graph AGI(R) with vertices V (AGI(R)) = {x ∈ R \ I : xy ∈ I for some y /∈ I}, and two distinct vertices x and y are adjacent if and only if AI(xy) 6= AI(x) ∪ AI(y). In other words, two distinct vertices x and y are adjacent if and only if there exists r ∈ R \ I such that rxy ∈ I and rx, ry /∈ I. Note that ΓI(R) is a subgraph of AGI(R). Also if I = {0}, then we have AGI(R) = AG(R). We call the graph AGI(R) the annihilator ideal graph.
In Section 2 of this paper, we study some basic properties of AGI(R). For instance, we show that if AGI(R) is not identical to ΓI(R), then the girth of the graph AGI(R) is at most 4 (see Theorem 2.7). In the third section, we study the planarity of AGI(R). In Section 4, we obtain some results on the annihilator ideal graph of Zn, where Znis the ring of integers modulo n for a positive integer n. We also study situations under which the graphs AGI(Zn) and ΓI(Zn) are isomorphic.
Now, we give a brief necessary background of graph theory. Let X be an undirected graph. We use the notation V (X) for the set of vertices of X. Also, for a vertex x ∈ V (X), N (x) denotes the set of vertices adjacent to x, and
|N (X)| is called the degree of x. We say that the graph X is connected if there is a path between each pair of distinct vertices of X. For two distinct vertices x and y, we define d(x, y) to be the length of a shortest path between x and y (d(x, y) = ∞ if there is no such path). The diameter of X is diam(X) = sup{d(x, y) : x and y are distinct vertices of X}. The girth of X is the length
of a shortest cycle in X, denoted by gr(X) ( gr(X) = ∞ if X has no cycles).
A clique of a graph is any complete subgraph of the graph and the number of vertices in a largest clique of X, denoted by ω(X), is called the clique number of X. A planar graph is a graph that can be embedded in the plane, that is, it can be drawn in the plane in such a way that its edges intersect only at their endpoints. Kuratowski provided a characterization of planar graphs, which is now known as Kuratowski’s Theorem: A finite graph is planar if and only if it does not contain a subdivision of K5 or K3,3 [6, Theorem 9.10]. Suppose that X is a graph with n vertices and q edges. Also, assume that C is a cycle of X. A chord in X is any edge of X joining two non-adjacent vertices in C. A primitive cycle is a cycle without chords. Moreover, we say that a graph X has the primitive cycle property (PCP) if any two primitive cycles intersect in at most one edge. The free rank of X, denoted by frank(X), is the number of primitive cycles of X. Also, the number rank(X) = q − n + r, where r is the number of connected components of X, is called the cycle rank of X. The cycle rank of X can be expressed as the dimension of the cycle space of X. These two numbers satisfy the inequality rank(X) ≤ frank(X), as is seen in [7, Proposition 2.2]. The precise definition of a ring graph can be found in Section 2 of [7]; the authors showed that, for the graph X, the following conditions are equivalent:
(i) X is a ring graph, (ii) rank(X) = frank(X),
(iii) X satisfies PCP and X does not contain a subdivision of K4as a subgraph.
Clearly ring graphs are planar. An undirected graph is an outerplanar graph if it can be drawn in the plane without crossings in such a way that all of the vertices belong to the unbounded face of the drawing. There is a characterization of outerplanar graphs that says a graph is outerplanar if and only if it does not contain a subdivision of the complete graph K4 or the complete bipartite graph K2,3 [8, Theorem 11.10]. Clearly, every outerplanar graph is planar. Let G and H be graphs. We use the notations G = H and G ∼= H to denote identical and isomorphic graphs, respectively.
As usual, Z and Zn will denote the rings of integers and integers modulo n, respectively.
1 Basic properties of AGI(R)
In this section, we investigate the basic properties of AGI(R). Also, we study the relations between the graphs AGI(R) and AG(R/I). The following proposition immediately follows from the definition of the graph AGI(R).
Proposition 1.1. Let I be a proper ideal of R. Then AGI(R) = ∅ if and only if I is a prime ideal.
Proposition 1.2. Let x + I and y + I be distinct elements in R/I. Then x + I is adjacent to y + I in AG(R/I) if and only if x is adjacent to y in AGI(R).
Proof. Since AI(xy) 6= AI(x)∪AI(y) if and only if annR/I(xy +I) 6= annR/I(x+
I) ∪ annR/I(y + I), we have x + I is adjacent to y + I in AG(R/I) if and only
if x is adjacent to y in AGI(R). QED
Theorem 1.3. If x + i is adjacent to y + i in AGI(R), for some i ∈ I, then all elements of x + I and y + Iare adjacent in AGI(R).
Proof. Assume that x + i is adjacent to y + i, for some i ∈ I. Hence there exists r ∈ R \ I such that r(x + i)(y + i) ∈ I, r(x + i) /∈ I, and r(y + i) /∈ I. Then, for every j, k ∈ I, we have r(x + j)(y + k) ∈ I, r(x + j) /∈ I, and r(y + k) /∈ I. So all vertices of x + I and y + I are adjacent in AGI(R). QED
In the following example, we show that if AG(R/I) ∼= AG(S/J ), where I and J are ideals of the rings R and S, respectively, then the graphs AGI(R) and AGJ(S) are not necessarily isomorphic.
Example 1.4. Let R = Z6× Z3 and I = 0 × Z3. Then AG(R/I) ∼= AG(Z6) with vertex-set {(2, 0), (3, 0), (4, 0)}, where (3, 0) is adjacent to both vertices (2, 0) and (4, 0). Let S = Z24 and J =< 8 >. Then AG(S/J ) ∼= AG(Z8) with vertex-set {2, 4, 6}, where the vertex 4 is adjacent to the vertices 2 and 6. Now the graph AGI(R) is pictured in Figure 1 while AGJ(S) is a complete graph.
Hence AG(R/I) ∼= AG(S/J ), but the graphs AGI(R) and AGJ(S) are not isomorphic.
(2,0) (3,0)
(4,0)
(2,1) (3,1)
(4,1)
(2,2)
(3,2)
(4,2)
Figure 1
Let {aλ}λ∈Λ⊆ R be a set of coset representatives of the vertices of AG(R/I).
In [13], S. P. Redmond defined the concept of a column and a connected column.
Here, we call the subset aλ+ I = {aλ+ i : i ∈ I} a column of AGI(R). Moreover, if there exists r ∈ R \ I such that raλ ∈ I and ra/ 2λ ∈ I, then we say that aλ+ I is a connected column of AGI(R).
Now the following two theorems follow immediately from the previous para- graph.
Theorem 1.5. If a + I is a connected column of AGI(R), then a + I is a complete subgraph of AGI(R) and ω(AGI(R)) ≥ |I|.
Theorem 1.6. If AGI(R) has a connected column and |V (AG(R/I))| ≥ 2, then ω(AGI(R)) ≥ |I| + 1.
In the next theorem, we study the girth of the graph AGI(R).
Theorem 1.7. Let AGI(R) 6= ΓI(R). Then gr(AGI(R)) ≤ 4.
Proof. Since AGI(R) 6= ΓI(R), there exist adjacent vertices x and y such that xy /∈ I. Thus there exists 1 6= r ∈ R \ I such that rxy ∈ I, rx /∈ I and ry /∈ I.
Now, if r /∈ {x, y}, then it is easy to see that r is a vertex of AGI(R) which is adjacent to both vertices x and y, and so we have the cycle x − r − y − x.
Without loss of generality, we may assume that r = x. Then clearly x 6= x2, y 6= xy and we have the cycle x − y − x2− xy − x. So the result holds. QED
2 Planar, outerplanar and ring graph annihilator ideal graphs
We begin this section with the following theorem.
Theorem 2.1. Let |V (AG(R/I))| = 1. Then AGI(R) is planar if and only if |I| 6 4.
Proof. Suppose that V (AG(R/I)) = {x + I}. Clearly x2 ∈ I and x + I is a connected column of AGI(R). Hence, by Theorem 1.5, AGI(R) is isomorphic to the complete graph K|I|. Therefore AGI(R) is planar if and only if |I| ≤
4. QED
Lemma 2.2. Let |V (AG(R/I))| > 2 and |I| ≥ 3. Then AGI(R) is not planar.
Proof. Suppose that a+I is a vertex of AG(R/I). Note that AG(R) is connected since Γ(R) is connected [2, Theorem 2.3], and Γ(R) is a subgraph of AG(R) with V (Γ(R)) = V (AG(R)). Now since |V (AG(R/I))| > 2 and AG(R/I) is connected, there exists b ∈ R \ I such that the vertices a and b are adjacent
in AGI(R). Hence AGI(R) has a subgraph isomorphic to K3,3 with vertex-set {a, a + x1, a + x2} ∪ {b, b + x1, b + x2}, where x1 and x2 are distinct elements in I \ {0}. So AGI(R) contains a subgraph which is isomorphic to K3,3. Thus
AGI(R) is not planar. QED
Theorem 2.3. Let |V (AG(R/I))| = 3 and |I| = 2. Then AGI(R) is not planar if and only if AG(R/I) is complete such that at least one of its vertices is a connected column in AGI(R).
Proof. First assume that AG(R/I) is a complete graph with vertices a + I, b + I and c + I. Without loss of generality, we may assume that a + I is a connected column. Then AGI(R) has a subgraph isomorphism to K3,3 with vertex-set {b, b + x, a} ∪ {c, c + x, a + x}, where x ∈ I \ {0}. Hence AGI(R) contains a copy of K3,3, which means that it is not planar.
For the converse statement, assume to the contrary that if AG(R/I) is com- plete, then none of its vertices is a connected column, or AG(R/I) is not com- plete. If AG(R/I) is complete and no vertex is a connected column, then AGI(R) is pictured in Figure 2, which is planar, and this is a contradiction.
b + x
a + x
c
c + x b
a Figure 2
Now assume that AG(R/I) is not a triangle, say that a + I and c + I are not adjacent. Then AGI(R) is pictured in Figure 3, which is planar, and this is a contradiction.
Hence the result holds. QED
Theorem 2.4. Let |V (AG(R/I))| > 4 and |I| = 2. Then AGI(R) is not planar if and only if there exists a subdivision of the complete graph K3 in AG(R/I).
Proof. First assume that there exists a subdivision of the complete graph K3 in AG(R/I). Suppose that a + I, b + I, c + I and d + I are distinct vertices in AG(R/I). Now if these vertices form a square in AG(R/I), then AGI(R) has a
a + x
b + x
c c + x
b a
Figure 3
subgraph isomorphic to K3,3 with vertex-set {a, a + x, c} ∪ {b, b + x, d}, where x ∈ I \ {0}. Hence AGI(R) contains a subgraph isomorphic to K3,3. So AGI(R) is not planar. If a, b and c form a triangle in AGI(R/I), then, in view of Figure 4, the graph AGI(R) contains a subdivision of K3,3. So AGI(R) is not planar.
a a +x
c + x
b c
d
b + x
Figure 4
The converse statement is clear. QED
Theorem 2.5. Let |V (AG(R/I))| = 1. Then |I| ≤ 3 if and only if AGI(R) is a ring graph and an outerplanar graph.
Proof. Since |V (A(R/I))| = 1, then, in view of the proof of Theorem 2.1, AGI(R) is isomorphic to K|I|. If |I| > 4, then AGI(R) contains a subgraph which is isomorphic to K4. So AGI(R) is neither a ring graph nor an outerpla- nar graph.
The converse statement is clear. QED
Since the graph AGI(R) is isomorphic to AG(R) whenever |I| = 1, in the following theorem, we verify the case that |I| > 2 and |V (A(R/I))| ≥ 2.
Theorem 2.6. Assume that |I|, |V (AG(R/I))| ≥ 2, and at least one of the sets V (AG(R/I)) or I has three elements. Then AGI(R) is neither outerplanar nor a ring graph.
Proof. First, suppose that |I| ≥ 3. Then the set of vertices {a, a + x1, a + x2} ∪ {b, b+x1}, where x1and x2are distinct nonzero elements of I and a, b are distinct elements of R \ I, forms the graph K2,3. So AGI(R) contains a subgraph which is isomorphic to K2,3. Assume that |V (AG(R/I))| ≥ 3 and that a + I, b + I, c + I are distinct vertices of AG(R/I). Since AG(R/I) is connected, we may assume that a + I and c + I are adjacent to b + I. Thus the vertices a and c are adjacent to b in AGI(R). Therefore the set of vertices {a, a + x1, c} ∪ {b, b + x1} forms the graph K2,3. Hence AGI(R) contains a subgraph which is isomorphic to K2,3. So AGI(R) is neither outerplanar nor a ring graph. QED
In the following theorem, we study the outerplanar and ring graph annihi- lator ideal graphs in the remaining case |I| = 2 and |V (A(R/I))| = 2.
Theorem 2.7. Let |I| = |V (AG(R/I))| = 2. Then AGI(R) is neither out- erplanar nor a ring graph if and only if there exist two connected columns in AGI(R).
Proof. Note that AGI(R) is connected since ΓI(R) is connected [13, Theorem 2.4], and ΓI(R) is a subgraph of AGI(R) with V (ΓI(R)) = V (AGI(R)). We have
|I| = |V (AG(R/I))| = 2, and |V (AGI(R))| = 4. So it is easy to see that there exist two connected columns in AGI(R) if and only if AGI(R) is isomorphic to
K4. Therefore the result holds. QED
3 Annihilator ideal graph of Zn
In this section, we assume that n is a positive integer and p, q are distinct prime numbers that divide n. Also, for x ∈ Zn, < x > denotes the ideal gener- ated by x in Zn.
Theorem 3.1. Let R = Zn and I =< pq >. Then AGI(Zn) = ΓI(Zn).
Proof. Let x and y be adjacent vertices of AGI(Zn). Then there exists r ∈ R \ I such that rxy ∈ I, rx ∈ R \ I, and ry ∈ R \ I. It is enough to show that xy ∈ I.
Since rxy ∈ I and p | pq, we have p | rxy, which implies that p | r, p | x or p | y, and q | r, q | x or q | y. Let p | r. Since pq - rx and pq - ry, we have q | r. Thus pq | r2, and this implies that pq | r. So r ∈ I, which is a contradiction. Let p | x.
Then q | y. So xy ∈ I. The case that p | y is obtained in a similar way. QED Theorem 3.2. Let R = Zn and I =< p2 >. Then AGI(Zn) = ΓI(Zn).
Proof. The proof is similar to the proof of Theorem 3.1. QED Theorem 3.3. Let R = Zpn and I =< pn−i >, for some 1 ≤ i ≤ n − 1.
Then AGI(R) is a complete graph.
Proof. Let x and y be distinct vertices of V (AGI(Zpn)). Then xx0 ∈ I and yy0 ∈ I, for some vertices x0, y0 ∈ V (AI(Zpn)). Hence pn−i| xx0 and pn−i | yy0. Set α = Max{α0 ∈ N : pα0 | x} and β = Max{β0 ∈ N : pβ0 | y}. Then we have pα+β | xy. Now, if α + β > n − i, then xy ∈ I, which implies that x is adjacent to y. Let α + β < n − i and put r = pn−i−(α+β). This implies that rxy ∈ I, and we have that rx ∈ R \ I. Assume that rx ∈ I. Then pn−i| pn−i−(α+β)pαa, where there exists a ∈ Z such that pαa = x. Thus pβ | a, and hence pα+β | x.
Therefore we have α + β ≤ α which implies that β = 0 and pn−i | y0. This is a
contradiction. QED
Theorem 3.4. Suppose that R = Zn and I = hki, where k ∈ Zn. If there exist vertices a, b, k ∈ Zn such that gcd(a, k) = gcd(b, k), then deg(a) = deg(b) and N (a) \ {b} = N (b) \ {a}.
Proof. Let s 6= b be an adjacent vertex to a. Then there exists r ∈ R \ I such that rsa ∈ I, rs /∈ I, and ra /∈ I. Thus k | rsa, k - rs, and k - ra. In order to show that s ∈ N (b), it is enough to prove that k | rsb, and k - rb. Set d = gcd(a, k) = gcd(b, k). Then we have k
d | rsa
d, and so k
d | rs, k | rsd and k | rsb. Now, assume on the contrary that k | rb. Then k
d | rb
d, and so k d | r, k
d | ra
d, and k | ra, which is contradiction. Therefore s ∈ N (b) \ {a}, and so the
results hold. QED
Theorem 3.5. Suppose that R = Zn and I = hki, where k ∈ Zn. If there exists an integer d > 1 such that d2 | k, then gr(AGI(Zn)) = 3.
Proof. Consider two distinct integers s1, s2 such that gcd(s1, k) = gcd(s2, k) = 1. Then gcd(s1d, k) = gcd(s2d, k) = d and, by Theorem 3.4, N (s1d) \ {s2d} = N (s2d) \ {s1d}. Now it is easy to see that s1d is adjacent to s2d and there exist at least three vertices in AGI(Zn), which completes the proof. QED
Acknowledgements. The authors are deeply grateful to the referee for careful reading of the manuscript and helpful suggestions.
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